Problem: The equation of a circle $C$ is $x^2+y^2+12x-8y+48 = 0$. What is its center $(h, k)$ and its radius $r$ ?
To find the equation in standard form, complete the square. $(x^2+12x) + (y^2-8y) = -48$ $(x^2+12x+36) + (y^2-8y+16) = -48 + 36 + 16$ $(x+6)^{2} + (y-4)^{2} = 4 = 2^2$ Thus, $(h, k) = (-6, 4)$ and $r = 2$.